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Based on the reaction below. How many grams of water would be form from 0.33 moles of C4H10 reacting with an unlimited amount of oxygen gas? (3pts) C4H10 + 6O2 --> 4CO2 + 5H2O
Mole ratio C4H10:H2O= 1:5= 0.33×5= 1.65molesMolar mass of H2O= 18.02= 18.02×1.65=29.733g
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1)Prepare n-butanyl bromide from n-butane
2)Prepare Ethyl bromide from n-butane
3)Prepare 1-Butanol from n-butane
4)Prepare 2-Butene from n-butane
1) CH3-CH2-CH2-CH3 + Cl2 ---gt; CH3-CH2-CH2-CH2-Cl + HCl CH3-CH2-CH2-CH2-Cl + KOH (dry) --gt; CH3-CH2-CH2=CH2 CH3-CH2-CH2=CH2 + Br2/H2O2 --gt; CH3-CH2-CH2-CH2-Br CH3-CH2-CH2-CH2-Br + KOH (aq) --gt; CH3-CH2-CH2-CH2-OH + KBr CH3-CH2-CH2-CH2-OH + NaBr/H2SO4...
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Butane is used as a fuel in camping gas stoves.100 cm3 of butane is burned in excess oxygen .calculate the volume of carbon dioxide produced at r.t.p
2 C4H10 + 13 O2 = 8 CO2 + 10 H2O n = V/22.4 n (C4H10) = 0.1/22.4 = 0.0047 mol n (CO2) = 4 · n (C4H10) = 4...
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If 5 moles of N<sub>2 </sub>reacted with 12 moles of O<sub>2 </sub>according to 0 N2 + 2 O<sub>2 </sub>+ NO<sub>2 </sub>how many moles NO<sub>2 </sub>are produced
Solution:Balanced chemical equation:N2 + 2O2 → 2NO2According to the equation above: 2×n(N2) = n(O2) = n(NO2)Moles of each reactant:n(N2) = 5 moln(O2) = 12 molChoose one reactant and determine how many...
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0.5 moles of carbon is burned completely in oxygen. How many moles of carbon dioxide are produced?
Solution:Balanced chemical equation:C(s) + O2(g) → CO2(g)According to stoichiometry:1 mol of C produces 1 mol of CO2Thus, 0.5 mol of C produces:(0.5 mol C) × (1 mol CO2 / 1 mol C)...
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Calculate the mass (in grams) of H2O produced from the reaction of 56.20 g C4H10 with 85.30 g O2.
C4H10 (g) + O2 (g) → CO2 (g) + H2O (l)
C4H10(g) + 6,5O2(g) → 4CO2(g) + 5H2O(l)n(C4H10)=m(C4H10)/M(C4H10)=56.20/58=0.969moln(O2)=m(O2)/M(O2)=85.30/32=2.666moln(C4H10)/1 and n(O2)/6.5 0.969/1 and 2.666/6.5 0.969 and 0.43The mass of water is calculated by the deficiency, by n(O2)n(O2)/n(H2O)=6.5/52.666*5=6.5*n(H2O)n(H2O)=2.05molV(H2O)=n(H2O)*Vm=2.05*22.4=45.92l=45920mlρ(H2O)=1g/mlm(H2O)=ρ*V(H2O)=1*45920=45920gAnswer - 45920g
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The density of butane, C4H10 , is 2.49×10−3 g/mL . Calculate the number of moles of butane in a 35.9 mL sample.
2.49×10−3 g/mL × 35.9 mL = 0.089391 g Butane molar mass 58.124 g·mol−1 0.089391 / 58.124 g·mol−1 = 0.001538 mol
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The complete combustion of heptane (C7H16) in oxygen produces carbon dioxide and water.
If I had 4.75g of heptane and combusted this to completion, how many grams of carbon dioxide would I produce?
100 gm of heptane will form 308 gm of carbon dioxide.So 4.75 gm of heptane will form 14.63 gm of carbon dioxide14.63 gm of CO2 will be produced
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2.00 grams of magnesium carbonate are reacted with excess hydrochloric acid. If the density of carbon dioxide gas at atmospheric conditions is 24 L/mol, what is the volume of the gas evolved?
The reaction equation:MgCO3 + 2HCl = MgCl2 + CO2 + H2OThe quantity of magnesium carbonate:2.00 g / (84.31 g/mol) = 0.0237 molThe quantity carbon dioxide is the same (according to...
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2 FeCl2 + Cl2 → 2 FeCl3 If 2.4154 moles of FeCl2 are reacted, how many moles of Cl2 will also be reacted?
Solution:The balanced chemical equation:2FeCl2 + Cl2 → 2FeCl3According to the equation above: n(FeCl2)/2 = n(Cl2)Hence,n(Cl2) = n(FeCl2)/2 = 2.4154 mol / 2 = 1.2077 moln(Cl2) = 1.2077 molAnswer: 1.2077 moles of Cl2 will also be...
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